Vertices of equilateral triangle inscribed in the unit circle

Multiplying through by $z_1^{-1}$ will rotate the numbers without changing any angles or distances, so you may assume $z_1=1$. Then $z_2$ and $z_3$ must have cancelling imaginary parts but can't be additive inverses; since they are on the unit circle, this forces them to be conjugate. So, they have the same real part, which must be $-1/2$ to cancel $z_1$. But you know where the numbers are on the unit circle that have real part $-1/2$.

$|z_1+z_2|^2=|z_3|^2\Rightarrow z_1\bar{z_2}+z_2\bar{z_1}=-1\Rightarrow|z_1|^2+|z_2|^2-z_1\bar{z_2}-z_2\bar{z_1}=|z_1-z_2|^2=3$

So $|z_1-z_2|=\sqrt{3}$ and similarly you have $|z_2-z_3|=|z_3-z_1|=\sqrt{3}.$